3.46 \(\int (c+d x)^{3/2} \sin ^2(a+b x) \, dx\)

Optimal. Leaf size=203 \[ \frac{3 \sqrt{\pi } d^{3/2} \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{32 b^{5/2}}-\frac{3 \sqrt{\pi } d^{3/2} \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{32 b^{5/2}}+\frac{3 d \sqrt{c+d x} \sin ^2(a+b x)}{8 b^2}-\frac{(c+d x)^{3/2} \sin (a+b x) \cos (a+b x)}{2 b}-\frac{3 d \sqrt{c+d x}}{16 b^2}+\frac{(c+d x)^{5/2}}{5 d} \]

[Out]

(-3*d*Sqrt[c + d*x])/(16*b^2) + (c + d*x)^(5/2)/(5*d) + (3*d^(3/2)*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*S
qrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(32*b^(5/2)) - (3*d^(3/2)*Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x
])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(32*b^(5/2)) - ((c + d*x)^(3/2)*Cos[a + b*x]*Sin[a + b*x])/(2*b)
+ (3*d*Sqrt[c + d*x]*Sin[a + b*x]^2)/(8*b^2)

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Rubi [A]  time = 0.359744, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3311, 32, 3312, 3306, 3305, 3351, 3304, 3352} \[ \frac{3 \sqrt{\pi } d^{3/2} \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{32 b^{5/2}}-\frac{3 \sqrt{\pi } d^{3/2} \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{32 b^{5/2}}+\frac{3 d \sqrt{c+d x} \sin ^2(a+b x)}{8 b^2}-\frac{(c+d x)^{3/2} \sin (a+b x) \cos (a+b x)}{2 b}-\frac{3 d \sqrt{c+d x}}{16 b^2}+\frac{(c+d x)^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(3/2)*Sin[a + b*x]^2,x]

[Out]

(-3*d*Sqrt[c + d*x])/(16*b^2) + (c + d*x)^(5/2)/(5*d) + (3*d^(3/2)*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*S
qrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(32*b^(5/2)) - (3*d^(3/2)*Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x
])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(32*b^(5/2)) - ((c + d*x)^(3/2)*Cos[a + b*x]*Sin[a + b*x])/(2*b)
+ (3*d*Sqrt[c + d*x]*Sin[a + b*x]^2)/(8*b^2)

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3306

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d +
f*x]/Sqrt[c + d*x], x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c
, d, e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(f*x^2)/d], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[(f*x^2)/d],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin{align*} \int (c+d x)^{3/2} \sin ^2(a+b x) \, dx &=-\frac{(c+d x)^{3/2} \cos (a+b x) \sin (a+b x)}{2 b}+\frac{3 d \sqrt{c+d x} \sin ^2(a+b x)}{8 b^2}+\frac{1}{2} \int (c+d x)^{3/2} \, dx-\frac{\left (3 d^2\right ) \int \frac{\sin ^2(a+b x)}{\sqrt{c+d x}} \, dx}{16 b^2}\\ &=\frac{(c+d x)^{5/2}}{5 d}-\frac{(c+d x)^{3/2} \cos (a+b x) \sin (a+b x)}{2 b}+\frac{3 d \sqrt{c+d x} \sin ^2(a+b x)}{8 b^2}-\frac{\left (3 d^2\right ) \int \left (\frac{1}{2 \sqrt{c+d x}}-\frac{\cos (2 a+2 b x)}{2 \sqrt{c+d x}}\right ) \, dx}{16 b^2}\\ &=-\frac{3 d \sqrt{c+d x}}{16 b^2}+\frac{(c+d x)^{5/2}}{5 d}-\frac{(c+d x)^{3/2} \cos (a+b x) \sin (a+b x)}{2 b}+\frac{3 d \sqrt{c+d x} \sin ^2(a+b x)}{8 b^2}+\frac{\left (3 d^2\right ) \int \frac{\cos (2 a+2 b x)}{\sqrt{c+d x}} \, dx}{32 b^2}\\ &=-\frac{3 d \sqrt{c+d x}}{16 b^2}+\frac{(c+d x)^{5/2}}{5 d}-\frac{(c+d x)^{3/2} \cos (a+b x) \sin (a+b x)}{2 b}+\frac{3 d \sqrt{c+d x} \sin ^2(a+b x)}{8 b^2}+\frac{\left (3 d^2 \cos \left (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{2 b c}{d}+2 b x\right )}{\sqrt{c+d x}} \, dx}{32 b^2}-\frac{\left (3 d^2 \sin \left (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{2 b c}{d}+2 b x\right )}{\sqrt{c+d x}} \, dx}{32 b^2}\\ &=-\frac{3 d \sqrt{c+d x}}{16 b^2}+\frac{(c+d x)^{5/2}}{5 d}-\frac{(c+d x)^{3/2} \cos (a+b x) \sin (a+b x)}{2 b}+\frac{3 d \sqrt{c+d x} \sin ^2(a+b x)}{8 b^2}+\frac{\left (3 d \cos \left (2 a-\frac{2 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{2 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{16 b^2}-\frac{\left (3 d \sin \left (2 a-\frac{2 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{2 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{16 b^2}\\ &=-\frac{3 d \sqrt{c+d x}}{16 b^2}+\frac{(c+d x)^{5/2}}{5 d}+\frac{3 d^{3/2} \sqrt{\pi } \cos \left (2 a-\frac{2 b c}{d}\right ) C\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{32 b^{5/2}}-\frac{3 d^{3/2} \sqrt{\pi } S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right ) \sin \left (2 a-\frac{2 b c}{d}\right )}{32 b^{5/2}}-\frac{(c+d x)^{3/2} \cos (a+b x) \sin (a+b x)}{2 b}+\frac{3 d \sqrt{c+d x} \sin ^2(a+b x)}{8 b^2}\\ \end{align*}

Mathematica [A]  time = 1.69637, size = 175, normalized size = 0.86 \[ \frac{\sqrt{\frac{b}{d}} \left (15 \sqrt{\pi } d^2 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )-15 \sqrt{\pi } d^2 \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )+2 \sqrt{\frac{b}{d}} \sqrt{c+d x} \left (4 b (c+d x) (4 b (c+d x)-5 d \sin (2 (a+b x)))-15 d^2 \cos (2 (a+b x))\right )\right )}{160 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(3/2)*Sin[a + b*x]^2,x]

[Out]

(Sqrt[b/d]*(15*d^2*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]] - 15*d^2*Sqrt[
Pi]*FresnelS[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]]*Sin[2*a - (2*b*c)/d] + 2*Sqrt[b/d]*Sqrt[c + d*x]*(-15*d^2*C
os[2*(a + b*x)] + 4*b*(c + d*x)*(4*b*(c + d*x) - 5*d*Sin[2*(a + b*x)]))))/(160*b^3)

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Maple [A]  time = 0.013, size = 197, normalized size = 1. \begin{align*} 2\,{\frac{1}{d} \left ( 1/10\, \left ( dx+c \right ) ^{5/2}-1/8\,{\frac{d \left ( dx+c \right ) ^{3/2}}{b}\sin \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }+3/8\,{\frac{d}{b} \left ( -1/4\,{\frac{d\sqrt{dx+c}}{b}\cos \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{da-cb}{d}} \right ) }+1/8\,{\frac{d\sqrt{\pi }}{b} \left ( \cos \left ( 2\,{\frac{da-cb}{d}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{dx+c}b}{d\sqrt{\pi }}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) -\sin \left ( 2\,{\frac{da-cb}{d}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{dx+c}b}{d\sqrt{\pi }}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(3/2)*sin(b*x+a)^2,x)

[Out]

2/d*(1/10*(d*x+c)^(5/2)-1/8/b*d*(d*x+c)^(3/2)*sin(2/d*(d*x+c)*b+2*(a*d-b*c)/d)+3/8/b*d*(-1/4/b*d*(d*x+c)^(1/2)
*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)+1/8/b*d*Pi^(1/2)/(b/d)^(1/2)*(cos(2*(a*d-b*c)/d)*FresnelC(2/Pi^(1/2)/(b/d)^(
1/2)*(d*x+c)^(1/2)*b/d)-sin(2*(a*d-b*c)/d)*FresnelS(2/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d))))

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Maxima [C]  time = 1.82384, size = 899, normalized size = 4.43 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/1280*sqrt(2)*(128*sqrt(2)*(d*x + c)^(5/2)*b^2*sqrt(abs(b)/abs(d)) - 160*sqrt(2)*(d*x + c)^(3/2)*b*d*sqrt(abs
(b)/abs(d))*sin(2*((d*x + c)*b - b*c + a*d)/d) - 120*sqrt(2)*sqrt(d*x + c)*d^2*sqrt(abs(b)/abs(d))*cos(2*((d*x
+ c)*b - b*c + a*d)/d) + ((15*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*sqr
t(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(
0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d
^2))))*d^2*cos(-2*(b*c - a*d)/d) - (15*I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))
) + 15*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*sqrt(pi)*sin(1/4*pi + 1/
2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0,
d/sqrt(d^2))))*d^2*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2*I*b/d)) + ((15*sqrt(pi)*cos(1/4*pi + 1/2*ar
ctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sq
rt(d^2))) + 15*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*I*sqrt(pi)*sin(-1
/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^2*cos(-2*(b*c - a*d)/d) - (-15*I*sqrt(pi)*cos(1/4*
pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*ar
ctan2(0, d/sqrt(d^2))) + 15*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*sqrt(p
i)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^2*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c
)*sqrt(-2*I*b/d)))/(b^2*d*sqrt(abs(b)/abs(d)))

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Fricas [A]  time = 2.15745, size = 478, normalized size = 2.35 \begin{align*} \frac{15 \, \pi d^{3} \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{C}\left (2 \, \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) - 15 \, \pi d^{3} \sqrt{\frac{b}{\pi d}} \operatorname{S}\left (2 \, \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) + 2 \,{\left (16 \, b^{3} d^{2} x^{2} + 32 \, b^{3} c d x + 16 \, b^{3} c^{2} - 30 \, b d^{2} \cos \left (b x + a\right )^{2} + 15 \, b d^{2} - 40 \,{\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt{d x + c}}{160 \, b^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

1/160*(15*pi*d^3*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d))) - 15*pi*d^3*
sqrt(b/(pi*d))*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) + 2*(16*b^3*d^2*x^2 + 32*b^3*
c*d*x + 16*b^3*c^2 - 30*b*d^2*cos(b*x + a)^2 + 15*b*d^2 - 40*(b^2*d^2*x + b^2*c*d)*cos(b*x + a)*sin(b*x + a))*
sqrt(d*x + c))/(b^3*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{\frac{3}{2}} \sin ^{2}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(3/2)*sin(b*x+a)**2,x)

[Out]

Integral((c + d*x)**(3/2)*sin(a + b*x)**2, x)

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Giac [C]  time = 1.26014, size = 772, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)*sin(b*x+a)^2,x, algorithm="giac")

[Out]

1/960*(192*(d*x + c)^(5/2) - 320*(d*x + c)^(3/2)*c - 20*(3*I*sqrt(pi)*d^2*erf(-sqrt(b*d)*sqrt(d*x + c)*(I*b*d/
sqrt(b^2*d^2) + 1)/d)*e^((2*I*b*c - 2*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) - 3*I*sqrt(pi)*d^2*erf
(-sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-2*I*b*c + 2*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2
*d^2) + 1)*b) - 16*(d*x + c)^(3/2) - 6*I*sqrt(d*x + c)*d*e^((2*I*(d*x + c)*b - 2*I*b*c + 2*I*a*d)/d)/b + 6*I*s
qrt(d*x + c)*d*e^((-2*I*(d*x + c)*b + 2*I*b*c - 2*I*a*d)/d)/b)*c + 15*sqrt(pi)*(4*I*b*c*d - 3*d^2)*d*erf(-sqrt
(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((2*I*b*c - 2*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1
)*b^2) + 15*sqrt(pi)*(-4*I*b*c*d - 3*d^2)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-2*
I*b*c + 2*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b^2) - 30*(-4*I*(d*x + c)^(3/2)*b*d + 4*I*sqrt(d*x +
c)*b*c*d + 3*sqrt(d*x + c)*d^2)*e^((2*I*(d*x + c)*b - 2*I*b*c + 2*I*a*d)/d)/b^2 - 30*(4*I*(d*x + c)^(3/2)*b*d
- 4*I*sqrt(d*x + c)*b*c*d + 3*sqrt(d*x + c)*d^2)*e^((-2*I*(d*x + c)*b + 2*I*b*c - 2*I*a*d)/d)/b^2)/d